What is the solution for xy^'-2y=-x^2-2x-1 ?

The solution for xy^'-2y=-x^2-2x-1 is y=-x^2ln(x)+2x+1/2+c_{1}x^2

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xy^'-2y=-x^2-2x-1

x y^{^{'}} - 2 y = - x^{2} - 2 x - 1

y = - x^{2} ln (x) + 2 x + \frac{1}{2} + c_{1} x^{2}

Solution steps

x y^{'} (x) - 2 y = - x^{2} - 2 x - 1

Solve linear ODE:

y = - x^{2} ln (x) + 2 x + \frac{1}{2} + c_{1} x^{2}

y = - x^{2} ln (x) + 2 x + \frac{1}{2} + c_{1} x^{2}

y^{^{'}} = x y^{- 2} - 2 y

x^{2} y^{^{'}} + x y = \frac{y ^{3}}{x}

e^{a r c t a n (4 t^{2})}

\frac{d}{d x} (- \frac{11}{x ^{5}})

\int x e^{6 x} d x

What is the solution for xy^'-2y=-x^2-2x-1 ?
The solution for xy^'-2y=-x^2-2x-1 is y=-x^2ln(x)+2x+1/2+c_{1}x^2